Brain Not Braining: Part 4 \ stacker news ~science

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433 sats \ 12 comments \ @noknees 11h science

I was stumped (but finally solved) by this seemingly simple question one of my friends challenged me, and I thought it would be good challenge for others here too :)

\sqrt{5+x} = 5 - x^2, \forall x \in \mathbb{R}

Find

x

for with proper steps/explanation for a 1000 sat bounty :)

There is a modern method to do it, but there's also a formula like the ol' quadratic formula for which I'll give extra sats if used :)

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1 reply \ @noknees OP 50m

I am designating @Scroogey as the winner for this bounty with a -200 sats (for admitting to use a solver lol, although that does not give away the answer).

He showed the right steps and right equations (check second pinned comment)!

Here is another simpler walkthrough by me though (simple factorization)

or using Ferrari's method for quartic polynomials:

0 sats \ 0 replies \ @SimpleStacker 18m

oh, i was gonna give it a go, but oh well haha

7 replies \ @Scroogey 7h

I would have squared both sides to

 $x^4 - 10x^2 - x + 20 = 0$

and then followed https://math.stackexchange.com/questions/785/general-formula-for-solving-quartic-degree-4-equations/1219804#1219804 with a=0, b=-10, c=-1, d=20. But the quadratic equation

 $v^2 + (2000-27-72*200)v + (100+12*20)^3 = 0$

has no solution, indicating that I made a mistake, and I grew tired :-)

0 sats \ 6 replies \ @noknees OP 7h

well that formula they showed is actually derived from the original quartic formula after depressing it and going case wise depending on the monic polynomial which is not the case for this one, this one is a quasi-symmetric variant, so it's not likely to give the answer :)

Good try tho! Most people won't reach till here...

400 sats \ 5 replies \ @Scroogey 6h

Oh, this is a better strategy: https://www.mathportal.org/formulas/algebra/solalgebric.php

With

a_1=0, a_2=-10, a_3=-1, a_4=20

the cubic equation becomes

 $y^3+10y^2-80y-801=0$

and with

y_1 = -9

(rational root test) the quadratic equations become

 $z^2+z-5=0$

 $z^2+z-4=0$

 $z^2-z-5=0$

 $z^2-z-4=0$

The solutions are

 $\frac{\pm 1 \pm \sqrt{21}}{2}$

and

 $\frac{\pm 1 \pm \sqrt{17}}{2}$

0 sats \ 2 replies \ @noknees OP 2h

How did you determine the coefficients so arbitrarily? The answer is correct though but I think that's cheating. Using factor theorem is a very bad way to solve this kind of equation. Anyways 200 sats for the good effort and somehow getting the correct answer :)

0 sats \ 1 reply \ @Scroogey 2h

I followed

from the link I posted.

0 sats \ 0 replies \ @noknees OP 1h

that's okay, but the rational root test link shows you solved the equation using a solver which is not okay, had you not posted the link and only written "rational root test", I might not have figured it out, but since you did, it puts a wrong impression

it's okay to use a solver once you understand the method, but citing is not always necessary! (another lesson I learnt the hard way)

but another 200 sats for being honest and using formulas the correct way!

0 sats \ 1 reply \ @Scroogey 6h

Since squaring both sides introduced false solutions, we have to check the four solutions above by inserting them into the original equation, which shows only two of them a true solutions:

 $\frac{1-\sqrt{21}}{2}$

and

 $\frac{-1+\sqrt{17}}{2}$